PROJECT MANAGEMENT: PERTjfratup.weebly.com/uploads/1/1/5/5/11551779/amat_167.15.pdf · BETA...

PROJECT MANAGEMENT: PERT

AMAT 167

PROBABILISTIC TIME ESTIMATES

We need three time estimates for each activity:

• Optimistic time (to): length of time required under optimum

conditions;

• Most likely time (tm): most probable amount of time required; and

• Pessimistic time (tp): length of time required under the worst

conditions;

BETA DISTRIBUTION

Confirming Pages

Chapter Seventeen Project Management 761

PROBABILISTIC TIME ESTIMATES The preceding discussion assumed that activity times were known and not subject to varia-tion. While that condition exists in some situations, there are many others where it does not. Consequently, those situations require a probabilistic approach.

The probabilistic approach involves three time estimates for each activity instead of one:

1. Optimistic time: The length of time required under optimum conditions; represented by t o .

2. Pessimistic time: The length of time required under the worst conditions; represented by t p .

3. Most likely time: The most probable amount of time required; represented by t m .

Managers or others with knowledge about the project can make these time estimates. The beta distribution is generally used to describe the inherent variability in time esti-

mates (see Figure 17.8 ). Although there is no real theoretical justification for using the beta distribution, it has certain features that make it attractive in practice: The distribution can be symmetrical or skewed to either the right or the left according to the nature of a particular activity; the mean and variance of the distribution can be readily obtained from the three time estimates listed above; and the distribution is unimodal with a high concentration of probabil-ity surrounding the most likely time estimate.

Of special interest in network analysis are the average or expected time for each activity, te, and the variance of each activity time, !i

2 . The expected time of an activity, te, is a weighted average of the three time estimates:

tt t t

eo m p

"# #4

6 (17–4)

Optimistic time The length of time required under optimal conditions.

Pessimistic time The length of time required under the worst conditions.

Most likely time The most probable length of time that will be required.

Beta distribution Used to describe the inherent variability in activity time estimates.





TABLE 17.3 Computer printout

SCHEDULE

EARLY LATEActivity Time ES EF LS LF Slack

1-2 8.00 0.00 8.00 0.00 8.00 0.001-3 4.00 0.00 4.00 6.00 10.00 6.002-4 6.00 8.00 14.00 10.00 16.00 2.002-5 11.00 8.00 19.00 8.00 19.00 0.003-5 9.00 4.00 13.00 10.00 19.00 6.004-5 3.00 14.00 17.00 16.00 19.00 2.005-6 1.00 19.00 20.00 19.00 20.00 0.00

THE CRITICAL PATH SEQUENCE IS:SNODE FNODE TIME

1 2 8.002 5 11.005 6 1.00

20.00

Activitystart

0

Optimistictime

Most likelytime (mode)

Pessimistictime

te tptmto

FIGURE 17.8 A beta distribution is used to describe probabilistic time estimates

ste25251_ch17_740-791.indd 761ste25251_ch17_740-791.indd 761 12/16/10 11:02:01 PM12/16/10 11:02:01 PM

This is used to describe the inherent variability in time estimates.

BETA DISTRIBUTION• There is no established theoretical justification for using the beta

distribution.

• It has certain features that make it attractive in practice:

ØThe distribution can be symmetrical or skewed to either the right or the left according to the nature of a particular activity;

Øthe mean and variance of the distribution can be readily obtained from the

three time estimates (to, tm, tp); and

Øthe distribution is unimodal with a high concentration of probability

surrounding the most likely time estimate.

AVERAGE TIME

Expected time of an activity:

Confirming Pages


PROBABILISTIC TIME ESTIMATES The preceding discussion assumed that activity times were known and not subject to varia-tion. While that condition exists in some situations, there are many others where it does not. Consequently, those situations require a probabilistic approach.

The probabilistic approach involves three time estimates for each activity instead of one:

1. Optimistic time: The length of time required under optimum conditions; represented by t o .

2. Pessimistic time: The length of time required under the worst conditions; represented by t p .

3. Most likely time: The most probable amount of time required; represented by t m .

Managers or others with knowledge about the project can make these time estimates. The beta distribution is generally used to describe the inherent variability in time esti-

mates (see Figure 17.8 ). Although there is no real theoretical justification for using the beta distribution, it has certain features that make it attractive in practice: The distribution can be symmetrical or skewed to either the right or the left according to the nature of a particular activity; the mean and variance of the distribution can be readily obtained from the three time estimates listed above; and the distribution is unimodal with a high concentration of probabil-ity surrounding the most likely time estimate.

Of special interest in network analysis are the average or expected time for each activity, te, and the variance of each activity time, !i

2 . The expected time of an activity, te, is a weighted average of the three time estimates:

tt t t

eo m p

"# #4

6 (17–4)









TABLE 17.3 Computer printout

SCHEDULE

EARLY LATEActivity Time ES EF LS LF Slack

1-2 8.00 0.00 8.00 0.00 8.00 0.001-3 4.00 0.00 4.00 6.00 10.00 6.002-4 6.00 8.00 14.00 10.00 16.00 2.002-5 11.00 8.00 19.00 8.00 19.00 0.003-5 9.00 4.00 13.00 10.00 19.00 6.004-5 3.00 14.00 17.00 16.00 19.00 2.005-6 1.00 19.00 20.00 19.00 20.00 0.00

THE CRITICAL PATH SEQUENCE IS:SNODE FNODE TIME

1 2 8.002 5 11.005 6 1.00

20.00

Activitystart

0

Optimistictime

Most likelytime (mode)

Pessimistictime

te tptmto

FIGURE 17.8 A beta distribution is used to describe probabilistic time estimates


Confirming Pages

762 Chapter Seventeen Project Management

The expected duration of a path (i.e., the path mean) is equal to the sum of the expected times of the activities on that path:

Path mean of expected times of activities! ∑ on the path (17–5)

The standard deviation of each activity’s time is estimated as one-sixth of the difference between the pessimistic and optimistic time estimates. (Analogously, nearly all of the area under a normal distribution lies within three standard deviations of the mean, which is a range of six standard deviations.) We find the variance by squaring the standard deviation. Thus,

" !# #2 or

( ) ( )t t t tp o p o

6 36

2 2⎡

⎣⎢

⎤

⎦⎥

(17–6)

The size of the variance reflects the degree of uncertainty associated with an activity’s time: The larger the variance, the greater the uncertainty.

It is also desirable to compute the standard deviation of the expected time for each path. We can do this by summing the variances of the activities on a path and then taking the square root of that number; that is,

" !path (Variances of activities on path)∑ (17–7)

Example 5 illustrates these computations.

The network diagram for a project is shown in the accompanying figure, with three time esti-mates for each activity. Activity times are in weeks. Do the following:

a. Compute the expected time for each activity and the expected duration for each path.

b. Identify the critical path.

c. Compute the variance of each activity and the variance and standard deviation of each path.

3-4-5d

3-5-7e

5-7-9f

2-4-6b

4-6-8h

1-3-4

a

2-3-6g

2-3-5c

3-4-6i

Most likelytime

Optimistictime

Pessimistictime

AOA diagram

3-4-5 3-5-7 5-7-9

AON diagram

d e f

1-3-4 2-4-6 2-3-5

a b c

2-3-6 4-6-8 3-4-6

g h i

FinishStart

E X A M P L E 5 e celx

www.mhhe.com/stevenson11e


Expected duration of a path:

VARIANCE

Note: the standard deviation of each activity’s time is estimated as one-sixth of the difference between the pessimistic and optimistic time estimates

Variance of an activity’s time:

Confirming Pages





" !# #2 or

( ) ( )t t t tp o p o

6 36

2 2⎡

⎣⎢

⎤

⎦⎥

(17–6)









3-4-5d

3-5-7e

5-7-9f

2-4-6b

4-6-8h

1-3-4

a

2-3-6g

2-3-5c

3-4-6

i

Most likelytime

Optimistictime

Pessimistictime

AOA diagram

3-4-5 3-5-7 5-7-9

AON diagram

d e f

1-3-4 2-4-6 2-3-5

a b c

2-3-6 4-6-8 3-4-6

g h i

FinishStart




The larger the variance, the greater the uncertainty.

Variance of a path’s time:

Confirming Pages





" !# #2 or

( ) ( )t t t tp o p o

6 36

2 2⎡

⎣⎢

⎤

⎦⎥

(17–6)









3-4-5d

3-5-7e

5-7-9f

2-4-6b

4-6-8h

1-3-4

a

2-3-6g

2-3-5c

3-4-6i

Most likelytime

Optimistictime

Pessimistictime

AOA diagram

3-4-5 3-5-7 5-7-9

AON diagram

d e f

1-3-4 2-4-6 2-3-5

a b c

2-3-6 4-6-8 3-4-6

g h i

FinishStart




EXAMPLE

Confirming Pages





" !# #2 or

( ) ( )t t t tp o p o

6 36

2 2⎡

⎣⎢

⎤

⎦⎥

(17–6)









3-4-5d

3-5-7e

5-7-9f

2-4-6b

4-6-8h

1-3-4a

2-3-6g

2-3-5c

3-4-6i

Most likelytime

Optimistictime

Pessimistictime

AOA diagram

3-4-5 3-5-7 5-7-9

AON diagram

d e f

1-3-4 2-4-6 2-3-5

a b c

2-3-6 4-6-8 3-4-6

g h i

FinishStart




EXAMPLE

Confirming Pages





" !# #2 or

( ) ( )t t t tp o p o

6 36

2 2⎡

⎣⎢

⎤

⎦⎥

(17–6)









3-4-5d

3-5-7e

5-7-9f

2-4-6b

4-6-8h

1-3-4

a

2-3-6g

2-3-5c

3-4-6i

Most likelytime

Optimistictime

Pessimistictime

AOA diagram

3-4-5 3-5-7 5-7-9

AON diagram

d e f

1-3-4 2-4-6 2-3-5

a b c

2-3-6 4-6-8 3-4-6

g h i

FinishStart




REMARK (TRIANGULAR DISTRIBUTION)

http://www.epixanalytics.com/modelassist/AtRisk/images/15/image219.gif

PATH PROBABILITY

Although activity times are represented by a beta distribution, the path distribution is represented by a normal distribution.

The central limit theorem tells us that the summing of activity times (random variables) results in a normal distribution. The normal tendency improves as the number of random variables increases. However, even when the number of items being summed is fairly small, the normal approximation provides a reasonable approximation to the actual distribution.

PATH PROBABILITY

Confirming Pages


Knowledge of the expected path times and their standard deviations enables a manager to compute probabilistic estimates of the project completion time, such as these:

The probability that the project will be completed by a specified time.

The probability that the project will take longer than its scheduled completion time.

These estimates can be derived from the probability that various paths will be completed by the specified time. This involves the use of the normal distribution. Although activity times are represented by a beta distribution, the path distribution is represented by a normal distribu-tion. The central limit theorem tells us that the summing of activity times (random variables) results in a normal distribution. This is illustrated in Figure 17.9 . The rationale for using a normal distribution is that sums of random variables (activity times) will tend to be normally distributed, regardless of the distributions of the variables. The normal tendency improves as the number of random variables increases. However, even when the number of items being summed is fairly small, the normal approximation provides a reasonable approximation to the actual distribution.

Normal

Path

Beta

Activity

!""Beta

Activity

Beta

Activity

FIGURE 17.9 Activity distributions and the path distribution

a.Path Activity

TIMESto tm tp t

t 4t t6e

o m p!

" "Path Total

a-b-c a 1 3 4 2.83b 2 4 6 4.00 10.00c 2 3 5 3.17

d-e-f d 3 4 5 4.00e 3 5 7 5.00 16.00f 5 7 9 7.00

g-h-i g 2 3 6 3.33h 4 6 8 6.00 13.50i 3 4 6 4.17

}}

}

b. The path that has the longest expected duration is the critical path. Because path d-e-f has the largest path total, it is the critical path.

c.

Path Activity

TIMES

to tm tps p o

act2

2(t t )36

!#

$path2 $path

a-b-c a 1 3 4 (4 ! 1)2/36 " 9/36b 2 4 6 (6 ! 2)2/36 " 16/36 34/36 " 0.944 0.97c 2 3 5 (5 ! 2)2/36 " 9/36

d-e-f d 3 4 5 (5 ! 3)2/36 " 4/36e 3 5 7 (7 ! 3)2/36 " 16/36 36/36 " 1.00 1.00f 5 7 9 (9 ! 5)2/36 " 16/36

g-h-i g 2 3 6 (6 ! 2)2/36 " 16/36h 4 6 8 (8 ! 4)2/36 " 16/36 41/36 " 1.139 1.07i 3 4 6 (6 ! 3)2/36 " 9/36

S O L U T I O N

}}

}


Confirming Pages


DETERMINING PATH PROBABILITIES The probability that a given path will be completed in a specified length of time can be deter-mined using the following formula:

z !"Specified time Path mean

Path standard deviiation (17–8)

The resulting value of z indicates how many standard deviations of the path distribution the specified time is beyond the expected path duration. The more positive the value, the better. (A negative value of z indicates that the specified time is earlier than the expected path dura-tion.) Once the value of z has been determined, it can be used to obtain the probability that the path will be completed by the specified time from Appendix B, Table B. Note that the probabil-ity is equal to the area under the normal curve to the left of z, as illustrated in Figure 17.10 .

If the value of z is # 3.00 or more, the path probability is close to 100 percent (for z ! # 3.00, it is .9987). Hence, it is very likely the activities that make up the path will be completed by the specified time. For that reason, a useful rule of thumb is to treat the path probability as being equal to 100 percent if the value of z is # 3.00 or more.

Rule of thumb: If the value of z is #3.00 or more, treat theprobability of path completion by the specified time as 100 percent.

0

Probabilityof completing

the path by thespecified time

Expected path duration

Specifiedtime

z

FIGURE 17.10 The path probability is the area under a normal curve to the left of z

A project is not completed until all of its activities have been completed, not only those on the critical path. It sometimes happens that another path ends up taking more time to complete than the critical path, in which case the project runs longer than expected. Hence, it can be risky to focus exclusively on the critical path. Instead, one must consider the possibility that at least one other path will delay timely project completion. This requires determining the prob-ability that all paths will finish by a specified time. To do that, find the probability that each path will finish by the specified time, and then multiply those probabilities. The result is the probability that the project will be completed by the specified time.

It is important to note the assumption of independence . It is assumed that path duration times are independent of each other. In essence, this requires two things: Activity times are independent of each other, and each activity is only on one path. For activity times to be independent, the time for one must not be a function of the time of another; if two activities were always early or late together, they would not be considered independent. The assumption of independent paths is usually considered to be met if only a few activities in a large project are on multiple paths. Even then, common sense should govern the decision of whether the independence assumption is justified.

Independence Assumption that path duration times are inde-pendent of each other; requiring that activity times be indepen-dent, and that each activity is on only one path.


Using the information from Example 5 , answer the following questions:

a. Can the paths be considered independent? Why?

b. What is the probability that the project can be completed within 17 weeks of its start?

c. What is the probability that the project will be completed within 15 weeks of its start?

d. What is the probability that the project will not be completed within 15 weeks of its start?




Probability that a given path will be completed in a specified length of time, can be determined using:

PATH PROBABILITY

Confirming Pages


DETERMINING PATH PROBABILITIES The probability that a given path will be completed in a specified length of time can be deter-mined using the following formula:

z !"Specified time Path mean

Path standard deviiation (17–8)

The resulting value of z indicates how many standard deviations of the path distribution the specified time is beyond the expected path duration. The more positive the value, the better. (A negative value of z indicates that the specified time is earlier than the expected path dura-tion.) Once the value of z has been determined, it can be used to obtain the probability that the path will be completed by the specified time from Appendix B, Table B. Note that the probabil-ity is equal to the area under the normal curve to the left of z, as illustrated in Figure 17.10 .

If the value of z is # 3.00 or more, the path probability is close to 100 percent (for z ! # 3.00, it is .9987). Hence, it is very likely the activities that make up the path will be completed by the specified time. For that reason, a useful rule of thumb is to treat the path probability as being equal to 100 percent if the value of z is # 3.00 or more.

Rule of thumb: If the value of z is #3.00 or more, treat theprobability of path completion by the specified time as 100 percent.

0

Probabilityof completing

the path by thespecified time

Expected path duration

Specifiedtime

z

FIGURE 17.10 The path probability is the area under a normal curve to the left of z

A project is not completed until all of its activities have been completed, not only those on the critical path. It sometimes happens that another path ends up taking more time to complete than the critical path, in which case the project runs longer than expected. Hence, it can be risky to focus exclusively on the critical path. Instead, one must consider the possibility that at least one other path will delay timely project completion. This requires determining the prob-ability that all paths will finish by a specified time. To do that, find the probability that each path will finish by the specified time, and then multiply those probabilities. The result is the probability that the project will be completed by the specified time.

It is important to note the assumption of independence . It is assumed that path duration times are independent of each other. In essence, this requires two things: Activity times are independent of each other, and each activity is only on one path. For activity times to be independent, the time for one must not be a function of the time of another; if two activities were always early or late together, they would not be considered independent. The assumption of independent paths is usually considered to be met if only a few activities in a large project are on multiple paths. Even then, common sense should govern the decision of whether the independence assumption is justified.



Using the information from Example 5 , answer the following questions:

a. Can the paths be considered independent? Why?

b. What is the probability that the project can be completed within 17 weeks of its start?

c. What is the probability that the project will be completed within 15 weeks of its start?

d. What is the probability that the project will not be completed within 15 weeks of its start?




PATH PROBABILITY

The resulting value of z indicates how many standard deviations of the path distribution the specified time is beyond the expected path duration. The more positive the value, the better.

Rule of thumb: treat the path probability as being equal to 100 percent if the value of z is +3.00 or more. Note: probability when z = +3.00 is .9987.

PATH PROBABILITY

A project is not completed until all of its activities have been completed, not only those on the critical path. It sometimes happens that another path ends up taking more time to complete than the critical path.

This requires determining the probability that all paths will finish by a specified time. To do that, find the probability that each path will finish by the specified time, and then multiply those probabilities. The result is the probability that the project will be completed by the specified time.

INDEPENDENCE

This assumes independence of each path duration times: “find the probability that each path will finish by the specified time, and then multiply those probabilities”.

To use this, activity times should be independent of each other, and each activity is only on one path.

However, this assumption is usually considered to be met if only a few activities in a large project are on multiple paths.

REMARK (NOT INDEPENDENT)In practice, when dependent cases occur, project planners often use

Monte Carlo Simulation.

EXERCISE

What is the probability that the project can be completed within 17 weeks of its start?

What is the probability that the project will not be completed within 17 weeks of its start?

Confirming Pages





" !# #2 or

( ) ( )t t t tp o p o

6 36

2 2⎡

⎣⎢

⎤

⎦⎥

(17–6)









3-4-5d

3-5-7e

5-7-9f

2-4-6b

4-6-8h

1-3-4

a

2-3-6g

2-3-5c

3-4-6

i

Most likelytime

Optimistictime

Pessimistictime

AOA diagram

3-4-5 3-5-7 5-7-9

AON diagram

d e f

1-3-4 2-4-6 2-3-5

a b c

2-3-6 4-6-8 3-4-6

g h i

FinishStart




SOLUTION

Confirming Pages


a. Yes, the paths can be considered independent, since no activity is on more than one path and you have no information suggesting that any activity times are interrelated.

b. To answer questions of this nature, you must take into account the degree to which the path distributions “overlap” the specified completion time. This overlap concept is illustrated in the accompanying figure, which shows the three path distributions, each centered on that path’s expected duration, and the specified completion time of 17 weeks. The shaded portion of each distribution corresponds to the probability that the part will be completed within the specified time. Observe that paths a-b-c and g-h-i are well enough to the left of the specified time, so that it is highly likely that both will be finished by week 17, but the critical path overlaps the specified completion time. In such cases, you need consider only the distribution of path d-e-f in assessing the probability of completion by week 17.

To find the probability for a path you must first compute the value of z using Formula 17–8 for the path. For example, for path d-e-f, we have

z !"

! #17 16

1 001 00

..

17weeks

Path

a-b-c

d-e-f

g-h-i

1.00

10.0

1.00

13.5

16.0

Weeks

Weeks

Weeks

Turning to Appendix B, Table B, with z ! # 1.00, you will find that the area under the curve to the left of z is .8413. The computations are summarized in the following table. Note: If the value of z exceeds # 3.00, treat the probability of completion as being equal to 1.000.

Pathz !

"17 ExpectedpathdurationPath standarddevviation

Probability of Completion in 17 Weeks

a-b-c 17 100 97

7 22"

! #.

.1.00

d-e-f 17 161 00

1 00"

! #.

..8413

g-h-i 17 13 51 07

3 27"

! #.

..

1.00

S O L U T I O N


SOLUTION

Confirming Pages


a. Yes, the paths can be considered independent, since no activity is on more than one path and you have no information suggesting that any activity times are interrelated.

b. To answer questions of this nature, you must take into account the degree to which the path distributions “overlap” the specified completion time. This overlap concept is illustrated in the accompanying figure, which shows the three path distributions, each centered on that path’s expected duration, and the specified completion time of 17 weeks. The shaded portion of each distribution corresponds to the probability that the part will be completed within the specified time. Observe that paths a-b-c and g-h-i are well enough to the left of the specified time, so that it is highly likely that both will be finished by week 17, but the critical path overlaps the specified completion time. In such cases, you need consider only the distribution of path d-e-f in assessing the probability of completion by week 17.

To find the probability for a path you must first compute the value of z using Formula 17–8 for the path. For example, for path d-e-f, we have

z !"

! #17 16

1 001 00

..

17weeks

Path

a-b-c

d-e-f

g-h-i

1.00

10.0

1.00

13.5

16.0

Weeks

Weeks

Weeks

Turning to Appendix B, Table B, with z ! # 1.00, you will find that the area under the curve to the left of z is .8413. The computations are summarized in the following table. Note: If the value of z exceeds # 3.00, treat the probability of completion as being equal to 1.000.

Pathz !



a-b-c 17 100 97

7 22"

! #.

.1.00

d-e-f 17 161 00

1 00"

! #.

..8413

g-h-i 17 13 51 07

3 27"

! #.

..

1.00

S O L U T I O N


Confirming Pages


P P P( ) ( )Finish by week 17 Path a-b-c finish! " (( ). .

(Path d-e-f finish Path g-"! " "1 00 8413

P hh-i finish). .1 00 8413!

c. For a specified time of 15 weeks, the z values are

Pathz !



a-b-c 15 10 0097

5 15#

! $.

..

1.00

d-e-f 15 16 001 00

1 00#

! #.

..

.1587

g-h-i 15 #! $

13 501 07

1 40.

..

.9192

Paths d-e-f and g-h-i have z values that are less than $ 3.00. From Appendix B, Table B, the area to the left of z ! # 1.00 is .1587, and the area to

the left of z ! $ 1.40 is .9192. The path distributions are illustrated in the figure. The joint probability of all finishing before week 15 is the product of their probabilities: 1.00(.1587)(.9192) ! .1459.

d. The probability of not finishing before week 15 is the complement of the probability obtained in part c: 1 # .1459 ! .8541.

15weeks

Path

a-b-c

d-e-f

g-h-i

1.00

10.0

.9192

13.5

16.0

Weeks

Weeks

Weeks

.1587

SIMULATION We have examined a method for computing the probability that a project would be completed in a specified length of time. That discussion assumed that the paths of the project were inde-pendent; that is, the same activities are not on more than one path. If an activity were on more than one path and it happened that the completion time for that activity far exceeded its expected time, all paths that included that activity would be affected and, hence, their times would not be independent. Where activities are on multiple paths, one must consider if the


PROJECT MANAGEMENT: PERTjfratup.weebly.com/uploads/1/1/5/5/11551779/amat_167.15.pdf · BETA...

Documents

Transcript of PROJECT MANAGEMENT: PERTjfratup.weebly.com/uploads/1/1/5/5/11551779/amat_167.15.pdf · BETA...